Free Objective Test 01 Practice Test - 11th and 12th

Question 1

An air bubble in sphere having 4 cm diameter appears 1 cm from surface nearest to eye when looked along diameter. If _am_g = 1.5, the distance of bubble from refracting surface is

[CPMT 2002]

1.2 cm

3.2 cm

2.8 cm

1.6 cm

SOLUTION

Solution : A

Question 2

A glass hemisphere of radius 0.04 m and R.I. of the material 1.6 is placed centrally over a cross mark on a paper (i) with the flat face; (ii) with the curved face in contact with the paper. In each case the cross mark is viewed directly from above. The position of the images will be

[ISM Dhanbad 1994]

(i) 0.04 m from the flat face; (ii) 0.025 m from the flat face

(i) At the same position of the cross mark; (ii) 0.025 m below the flat face

(i) 0.025 m from the flat face; (ii) 0.04 m from the flat face

For both (i) and (ii) 0.025 m from the highest point of the hemisphere

SOLUTION

Solution : B

Question 3

A converging lens is used to form an image on a screen. When upper half of the lens is covered by an opaque screen

A. Half the image will disappear

B. Complete image will be formed of same intensity

C. Half image will be formed of same intensity

D. Complete image will be formed of decreased intensity

SOLUTION

Solution : D

To form the complete image only two rays are to be passed through the lens. Moreover the total amount of light released by the object is not passing through the lens because of the covering. Therefore the intensity of the image will be reduced and the image formed will be faint, not sharp.

Question 4

A point object is placed at the center of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is

A. 2 cm

B. 4 cm

C. 6 cm

D. 12 cm

SOLUTION

Solution : C

Using refraction formula $\frac{_{1} μ_{2} - 1}{R} = \frac{_{1} μ_{2}}{v} - \frac{1}{u}$
in given case, medium (1) is glass and (2) is air
So $\frac{_{g} μ_{a} - 1}{R} = \frac{_{g} μ_{a}}{v} - \frac{1}{u} \Rightarrow \frac{\frac{1}{1.5} - 1}{- 6} = \frac{1}{1.5 v} - \frac{1}{- 6}$
$\Rightarrow \frac{1 - 1.5}{- 6} = \frac{1}{v} + \frac{1.5}{6} \Rightarrow \frac{0.5}{v} = \frac{1}{v} + \frac{1}{4}$
$\Rightarrow \frac{1}{v} = \frac{1}{12} - \frac{1}{4} = - \frac{2}{12} = - \frac{1}{6} \Rightarrow v = 6 c m .$

Question 5

Shown in the figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length + 20 cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is

A. + 80 cm

B. – 80 cm

C. – 24 cm

D. –100 cm

SOLUTION

Solution : D

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}$

$\frac{1}{f_{1}} = (1.6 - 1) (\frac{1}{\infty} - \frac{1}{20}) = \frac{0.6}{20} = - \frac{3}{100} . . . (i)$
$\frac{1}{f_{2}} = (1.5 - 1) (\frac{1}{20} - \frac{1}{- 20}) = \frac{1}{20} . . . (i i)$
$\frac{1}{f_{3}} = (1.6 - 1) (\frac{1}{- 20} - \frac{1}{\infty}) = - \frac{3}{100} . . . (i i i)$
$\Rightarrow \frac{1}{F} = - \frac{3}{100} + \frac{1}{20} - \frac{3}{100} \Rightarrow F = - 100_{c m}$

Question 6

A ray of light falls on a transparent sphere with centre at C as shown in figure. The ray emerges from the sphere parallel to line AB. The refractive index of the sphere is:

\sqrt{2}

\sqrt{3}

\frac{3}{2}

\frac{1}{2}

SOLUTION

Solution : B

Deviation by a aphere is 2(i -r) Here, deviation $δ = 60^{\circ} = 2 (i - r)$
(or) $i - r = 30^{\circ}$
$∴$ $r = i - 30^{\circ} = 60^{\circ} - 30^{\circ} = 30^{\circ}$
$∴$ $μ = \frac{s i n i}{s i n r} = \frac{s i n 60^{\circ}}{s i n 30^{\circ}} = \sqrt{3}$

Question 7

A transparent cylinder has its right half polished so as to act as a mirror. A paraxial light ray is incident from left, that is parallel to principal axis, exits parallel to the incident ray as shown. The refractive index of the material of the cylinder is:

A. 1.2

B. 1.5

C. 1.8

D. 2.0

SOLUTION

Solution : D

For spherical surface using $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$
$\frac{n}{2 R} - \frac{1}{\infty} = \frac{n - 1}{R}$ ; $n = 2 n - 2 \Rightarrow n = 2$ .

Question 8

Two identical glass $(μ_{g} = \frac{3}{2})$ equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water $(μ_{w} = \frac{4}{3})$ . The focal length of the combination is

f

\frac{f}{2}

\frac{4 f}{3}

\frac{3 f}{4}

SOLUTION

Solution : D

Let R be the radius of curvature of each surface. Then
$\frac{1}{f} = (1.5 - 1) (\frac{1}{R} + \frac{1}{R})$
$∴ R = f$
For the water lens
$\frac{1}{f} = (\frac{4}{3} - 1) (- \frac{1}{R} - \frac{1}{R}) = \frac{1}{3} (- \frac{2}{R})$
or $\frac{1}{f} = - \frac{2}{3 R}$
Now using $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}$ we have
$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} + \frac{1}{f}$
$= \frac{2}{f} - \frac{2}{3 f} = \frac{4}{3 f}$ $∴ F = \frac{3 f}{4}$

Question 9

A ray of light falls on the surface of a spherical glass paper weight making an angle $α$ with the normal and is refracted in the medium at an angle $β$ . The angle of deviation of the emergent ray from the direction of the incident ray is

(α - β)

2 (α - β)

\frac{(α - β)}{2}

(α + β)

SOLUTION

Solution : B

From figure it is clear that $Δ O B C$ is an isosceles triangle ,
Hence $∠ O C B = β$ and emergent angle is $α$
Also sum of two in terior angles = exterior angle
$∴ δ = (α - β) + (α - β) = 2 (α - β)$

Question 10

Refraction takes place at a convex spherical boundary separating air-glass medium. For the image to be real, the object distance $(μ_{g} = \frac{3}{2})$

A. Should be greater than two times the radius of curvature of the refracting surface

B. Should be greater than six times the radius of curvature of the refracting surface

C. Should be greater than the radius of curvature of the refracting surface

D. Is independent of the radius of curvature of the refracting surface

SOLUTION

Solution : A

Applying $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
$\frac{1.5}{v} - \frac{1}{- u} = \frac{(1.5) - 1}{R}$
Dividing with $1.5$ on both the sides.
or $\frac{1}{v} + \frac{2}{3 u} = \frac{1}{3 R}$
For v to be positive $\frac{1}{3 R} > \frac{2}{3 u}$
or $u > 2 R$

Question 11

A convex lens of crown glass ( =1.525) will behave as a divergent lens if immersed in

A. Water (n =1.33)

B. In a medium of n = 1.525

C. Carbon disulphide n =1.66

D. It cannot act as a divergent lens

SOLUTION

Solution : C

A lens shows opposite behaviour if $μ_{m e d i u m} > μ_{l e n s}$ .e., a convergent lens becomes divergent in nature

Question 12

A. + 80 cm

B. - 80 cm

C. -24 cm

D. -100 cm

SOLUTION

Solution : D

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}$

$\frac{1}{f_{1}} = (1.6 - 1) (\frac{1}{\infty} - \frac{1}{20}) = \frac{0.6}{20} = - \frac{3}{100} . . . (i) \frac{1}{f_{2}} = (1.5 - 1) (\frac{1}{20} - \frac{1}{- 20}) = \frac{1}{20} . . . (i i) \frac{1}{f_{3}} = (1.6 - 1) (\frac{1}{- 20} - \frac{1}{\infty}) = - \frac{3}{100} . . . . (i i i) \Rightarrow \frac{1}{F} = - \frac{3}{100} + \frac{1}{20} - \frac{3}{100} \Rightarrow F = - 100 c m$

Question 13

A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids $L_{1}$ and $L_{2}$ having refractive indices $n_{1}$ and $n_{2}$ respectively $(n_{2} > n_{1} > 1)$ . The lens will diverge a parallel beam of light if it is filled with

A. Air and placed in air

B. Air and immersed in

L_{i}

L_{1}

and immersed in

L_{2}

L_{2}

and immersed in

L_{1}

SOLUTION

Solution : D

$\frac{1}{f} = (\frac{n_{2}}{n_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ where $n_{2}$ and $n_{1}$ are the refractive indices of the material of the lens and of the surroundings respectively. For a double concave lens, $(\frac{1}{R_{1}} - \frac{1}{R_{2}})$ is always negative.

Hence f is negative only when $n_{2} > n_{1}$

Question 14

There is an equiconvex glass lens with radius of each face as R and $_{a} μ_{g} = \frac{3}{2} a n d_{a} μ_{w} = \frac{4}{3}$ . If there is water in object space and air in image space, then the focal length is

A. 2R

B. R

3 \frac{R}{2}

R^{2}

SOLUTION

Solution : C

Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium
$\frac{μ_{2} - μ_{1}}{R} = \frac{μ_{1}}{- u} + \frac{μ_{2}}{v_{1}} \Rightarrow \frac{(\frac{3}{2}) - (\frac{4}{3})}{R} = \frac{\frac{4}{3}}{\infty} + \frac{\frac{3}{2}}{v_{1}} \Rightarrow v_{1} = 9 R$
Now consider the refraction at the second surface of the lens i.e. refraction from denser medium to rarer medium
$\frac{1 - \frac{3}{2}}{- R} = - \frac{\frac{3}{2}}{9 R} + \frac{1}{v_{2}} \Rightarrow v_{2} = (\frac{3}{2}) R$
The image will be formed at a distance of $\frac{3}{2} R$ .This is equal to the focal length of the lens.

Question 15

A thin equiconvex lens is made of glass of R.I 1.5 and its focal length is 0.2 If it acts a concave lens of 0.5 m focal length when dipped in a liquid, the R.I of liquid is

\frac{17}{8}

\frac{15}{8}

\frac{13}{8}

\frac{9}{8}

SOLUTION

Solution : B

$- \frac{f_{1}}{f_{a}} = \frac{(μ_{5} - 1)}{(\frac{μ_{s}}{μ_{l}} - 1)} - 2.5 = \frac{0.5}{(\frac{1.5}{μ_{l}} - 1)} - \frac{0.5}{0.2} = \frac{1.5 - 1}{(\frac{1.5}{μ_{1}} - 1)} \frac{1.5}{μ_{l}} - 1 = - \frac{1}{5} \frac{1.5}{μ_{l}} = 1 - \frac{1}{5} = \frac{4}{5} μ_{l} = \frac{7}{4} = \frac{75}{40} = \frac{15}{8}$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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